* Giải hệ phương trình đối xứng loại I:
$\begin{cases}(x+y)\left ( 1+\frac{1}{xy} \right )=5 \\ (x^{2}+y^{2})\left ( 1+\frac{1}{x^{2}y^{2}} \right )=49 \end{cases}$
* Giải hệ phương trình đối xứng loại II:
Bài 1. $\begin{cases}2x+y=\frac{3}{x^{2}} \\ 2y+x=\frac{3}{y^{2}} \end{cases}$
Bài 2. $\begin{cases}3y=\frac{y^{2}+2}{x^{2}} \\ 3x=\frac{x^{2}+2}{y^{2}} \end{cases} \begin{cases}\left ( x, y\in R\right ) \end{cases}$
Bài 4. $\begin{cases}\sqrt{x^{2}+9}+y=9 \\ \sqrt{y^{2}+9}+x=9 \end{cases}$
2. Tu he co     $x,y>0     (1)$

$hpt<=>\begin{cases}3yx^2=x^2+2 \\ 3xy^2=y^2+2 \end{cases}$

      $=>3(x-y)(3xy+x+y)=0$

       $<=>x=y     do   (1)$

Tu do de dang co  $x=y=1$



4. Tru 2 ve pt co 

$\sqrt{x^2+9}-\sqrt{y^2+9}+y-x=0$

$<=>\frac{x^2-y^2}{\sqrt{x^2+9}+\sqrt{y^2+9}}+y-x=0$

$<=>(x-y)(\frac{x+y}{\sqrt{x^2+9}+\sqrt{y^2+9}-1})=0$

Vi   $x<\sqrt{x^2+9}       va      y<\sqrt{y^2+9}$

Nen    $x+y<\sqrt{x^2+9}+\sqrt{y^2+9}=>\frac{x+y}{\sqrt{x^2+9}+\sqrt{y^2+9}}<1$

Tu do suy ra  $x=y$

De dang tim dk   $x=y=4$
1.      $\begin{cases}2x+y=\frac{3}{x^2} \\ 2y+x=\frac{3}{y^2} \end{cases}$

Tru 2 ve cua pt cho nhau ta dk

        $(x-y)(1+\frac{3(x+y)}{x^2.y^2})=0$

$TH1:   x=y $  de dang giai dk $x=y=1$

$TH2:    1+\frac{3x+3y}{x^2y^2}=0<=>-3(x+y)=x^2y^2$

Cong hai ve cua hpt ban dau co 

           $3(x+y)=\frac{3x^2+3y^2}{x^2y^2}<=>3(x+y)=\frac{3x^2+3y^2}{-3(x+y)}$ 

$<=>-3(x+y)^2=x^2+y^2<=>2x^2+3xy+2y^2=0    (vo    nghiem)      $
Co    $(x+y)(1+\frac{1}{xy})=5$

        $<=>(x+\frac{1}{x})+(y+\frac{1}{y})=5$

Tuong tu co 

  $Pt2<=> x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=49$

Dat     $x+\frac{1}{x}=a$    va   $y+\frac{1}{y}=b$

Tu do ta co he 

           $\begin{cases}a+b=5 \\ a^2+b^2=53 \end{cases}$ 

De dang tim dk 
                        
              $\begin{cases}a=7 \\ b=-2 \end{cases}$ hoac   $\begin{cases}a=-2 \\ b= 7\end{cases}$

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